Let $X$ be a random variable with mean, $E(X)=\mu$ and variance, $E(X-\mu)^2=\sigma^2$. Then Chebyshev's inequality asserts that $$ P\{|X-\mu|\geq k\sigma\} \leq \frac{1}{k^2} $$

Using this inequality, I have to show that $$e^{k+1} \geq k^2 \,\,\text{for}\,\, k>1$$ It is clear that if I can show that $$P\{|X-\mu|\geq k\sigma\}\geq \frac{1}{e^{k+1}}$$ for $k>1$, and any $\mu \in \mathbb{R}$ and any $\sigma>0$, then we are done. Thanks in advance.

Source : Rohatgi, Saleh-p.$98$-problem $6$.


Consider a random variable $X$ that follows the exponential distribution $\mathsf{Exp}(1)$. We have $$ \mathsf{E}(X) = 1, \quad\mathsf{Var}(X) = 1 $$ and for $x > 0$, $$ \Pr(X \geq x) = e^{-x} $$ implying $\Pr(X \geq k + 1) = 1/ e^{k+1}$. In addition, by Chebyshev's bound, we have $$ \Pr(X \geq 1 + k) \leq 1/k^2 $$ Therefore, $$ 1 / e^{k+1} \leq 1/k^2 $$

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